# Schmitt Trigger: NAND Gates

Discussion in 'Homework Help' started by Steve1992, Oct 9, 2010.

1. ### Steve1992 Thread Starter Senior Member

Apr 7, 2006
100
0
When P is high, its voltage is 4.5 V.
M must go below 2.5 V to make P go low at 0.5 V.

Use calculations to show that L has to be taken below 1.1 V to make P go low.

My calculations:

4.5V (P)  1.1V (L) = 3.4V

3.4V/(330k + 470k) = 4.25uA

4.25uA*330k = 1.4V

1.1V (L) + 1.4V = 2.5V = M

Right?

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2. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
You cannot use 1.1V in your calculation because that is the voltage you are supposed to calculate.

Because of the extremely high input impedance of the CMOS gate, the input current is basically 0. Therefore your problem can be simplified to the finding of an input voltage that gives point 'M' 2.5V.

Try again.

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3. ### Steve1992 Thread Starter Senior Member

Apr 7, 2006
100
0
(4.5V - 2.5V)/470k = 4.25uA

2.5V/330k = 7.57uA

7.57uA - 4.25uA = 3.32uA

3.32uA*330k = 1.1V

4. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
Correct.

This is the only current in the loop. There is no other current path.

Wrong. The current in 330K is the same as 470K because they are in series. The current in 470K has already shown to be 4.25uA, so current in 330K is also 4.25uA.

Voltage drop across 330K is thus 4.25uA * 330K = 1.403V

With M = 2.5V, it follows that the input voltage must be 2.5 - 1.403 = 1.098V.

Apr 7, 2006
100
0
Thank you.